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Electric Circuits AC Series Circuits Presentation
- 1. Course Title: Electric Circuits (ECI:22324) Presentation on Unit 1: AC series circuits On Topics 1.1 Generation of alternating voltage, phasor representation of sinusoidal quantities
- 2. 1.1 Generation of alternating voltage and currents οAn alternating voltage may be generated: 1. By rotating a coil at constant angular velocity in a uniform magnetic field. 2. By rotating a magnetic field at a constant angular velocity within a stationary coil.
- 4. Continueβ¦. Consider a rectangular coil of n turns rotating in anticlockwise direction with an angular velocity of w rad/sec in uniform magnetic field. β’ Let the time be measured from the instant the plane of the coil coincide with OX-axis. In this position of the coil, the flux linking with the coil has its maximum value max. Fig.1.2 (i) β’ Let the coil turn through an angle ΞΈ(=Οt) in anticlockwise direction in t seconds and assume the position shown in fig 1.2(ii)flux . β’ In this position, the maximum flux max acting vertically downward can be resolved into two perpendicular components: β’ i) Component max sinΟt parallel to the plane the coil. This component induces no emf in the coil. β’ ii) Component max cosΟt perpendicular to the plane of the coil. This component induces emf in the coil.
- 5. Continueβ¦ β’ Flux linkage of the coil at the considered instant (i.e. at ΞΈ angle) =No. of turns * flux linking = n max cosΟt β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦β¦β¦.(1) Fig.
- 6. Continueβ¦. β’ According to Faradayβs law of electromagnetic induction, the emf induced in coil is equal to the rate of change of flux linkage of the coil. Hence emf v at the considered instant is given by, β΄ π£ = β π ππ‘ (π β max cosππ‘) β΄ π£ = βπβ maxπ(βsinππ‘) β΄ π£ = ππβ max(sinππ‘) β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(2) Value of π£ is maximum, when sinΟt=1 β΄ ππ = πβ maxΟ From equation (2), β΄ π£ = Vm(π ππΟt) β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(3) From equation (3), β’ If a coil rotating with a constant angular velocity in a uniform magnetic field produces a sinusoidal alternating emf. Similarly, equation of the alternating current is given by, β΄ π = ImsinΟt β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...(4)
- 7. 1.12 Important AC Terminology 1.) Waveform: The shape of the curve obtained by plotting the instantaneous value of voltage or current as ordinate against time is called waveform. 2.) Instantaneous Value: The value of alternating quantity at any instant is called instantaneous value. The instantaneous value of alternating voltage and current are represented by π£ and π . 3.) Cycle: One complete set of positive and negative values of an alternating quantity is known as cycle.
- 8. Continueβ¦. 4.) Alteration: One half cycle of an alternating quantity is called alteration. An alteration 180 degree electrical. 5.) Time period: The time taken in seconds to complete one cycle of an alternating quantity is called its time period. It is generally represented by π. 6.) Frequency: The number of cycle complete in one seconds is called the frequency (π)of the alternating quantity. It is measured in cycle/sec or hertz. One hertz is equal to 1cycles/seconds. 7.) Amplitude: The maximum value attained by an alternating quantity is called its amplitude or peak value. The amplitude of an alternating voltage or current is designated by π π or πΌπ respectively.
- 9. 1.13 Important Relations οTime period and frequency: β’ Consider an alternating quantity having a frequency of f c/s (Hz) and time period T sec. β’ Time taken to complete f cycle=1 seconds β’ Time taken to complete 1 cycle= 1/f second. β’ But the time taken to complete one cycle is the time period T β’ π = 1 π ππ π = 1 π οAngular velocity and frequency: β’ In a one revolution of the coil the angle turned is 2 radian and voltage wave cycle complete one cycle. β’ π΄πππ’πππ πππππππ‘π¦ = π΄ππππ π‘π’ππππ ππππ π‘ππππ β’ π = 2π π β’ π = 2π β 1 π β’ π = 2ππ
- 10. 1.14 Values of alternating voltage and current i. Peak Value ii. Average value or mean value iii. RMS value(root mean square) or effective value iv. Peak to peak value
- 11. Continue.. i. Peak Value: β’ It is maximum value attained by alternating quantity. β’ It is represented by π π and πΌπ of alternating voltage and current respectively. β’ Peak value is important in case of testing of material. β’ Peak value is not used to specify the magnitude of alternating voltage or current.
- 12. Continue⦠ii. Average value: ⒠The average value of alternating current is zero over one cycle because positive area exactly cancels the negative area. ⒠However, half cycles average value is not zero. ⒠Therefore , whenever the average value of alternating current or voltage is asked, it is understood for half cycles.
- 13. Continueβ¦ Expression for alternating current is π = π°πππππ½ β¦β¦β¦β¦β¦ΞΈ=Οt β’ Average value of current is given by π°ππ = ππππ ππ ππππ πππππ ππππ ππππππ ππ ππππ πππππ π°ππ = π π ππ π½ π = π π π π π°πππππ½π π½ = πΌπ π βπππ π 0 π = πΌπ π βπππ π β (βπππ 0) = πΌπ π β(β1) β (β 1 )
- 14. Continueβ¦ β’ Solving above integration we get, π°ππ = π π π°π = π. ππππ°π π΄ππππππ β’ Similarly, for alternating voltage varying sinusoidally, π½ππ = π π π½π = π. ππππ½π Volts.
- 15. Continueβ¦ iii. RMS (root mean square) value or effective value: β’ The equation of the alternating current varying sinusoidally is given by, π = πΌππ πππ β’ Consider an elementary strip of thickness ππ in first half cycle of the squared current wave. β’ let π2 be the mid ordinate of the strip β’ Therefore, ππππ ππ πππ πππππ = ππ π π½ Area of half squared wave = π π πππ π½ = π π π°π ππππππ½π π½ π
- 16. Continue.. β’ Solving above integration, we get Area of half squared wave= π°π π π π β’ Let, π°πππ = β ππππ ππ ππππ πππππ πππππππ ππππ ππππ ππππππ ππππ = β =π°π π π π π = π°π π π = π°π βπ = π. πππ π°π β’ Similarly, for alternating voltage varying sinusoidally, π½πππ = π½π βπ = π. πππ π½π β’ Note: π°πππ = π°, π½πππ = π½
- 17. Continueβ¦ iv. Peak to Peak value: (1) ππππ ππππππ = πΉπ΄πΊ πππππ π¨ππππππ πππππ = π.πππβπ¦ππ± πππππ π.πππβπππ.πππππ = π. ππ (2) Peak ππππππ = πππ. πππππ πΉπ΄πΊ πππππ = π¦ππ± πππππ π.πππβπππ.πππππ = π. πππ β’ The peak factor is great important because it indicates the maximum value voltage being applied to the various of the apparatus.
- 18. Continueβ¦ β’ Phase and phase difference: οPhase: phase of particular value of an alternating quantity is the fractional part of time period or cycle through which the quantity has advance from the selected zero position or reference. οPhase difference refers to the angular displacement between different waveforms of the same frequency.
- 19. Phasor representation of alternating quantities
- 20. Phasor representation of alternating quantities β’ When number of waveforms are drawn in the same figure, the complexity of diagram increases and it becomes very difficult to extract the information from the waveforms. Therefore, to extract the same information, simplified alternate approach is preferred, called βPhasor representation of Sinusoidal quantityβ. β’ A sinusoidal quantity is represented by a rotating vector or rotating phasor βAβ whose length is equal to the amplitude of the quantity βAmβ, as shown above. The points on the waveform are represented by the positions of the phasor during rotation drawn from the same reference point. The phasor making an angle of ππ‘ with respect to positive x-axis reference, represents the instantaneous value of the quantity at an angle of ππ‘ from its zero value, as shown above. In fact, the vertical component of the phasor represents the magnitude of the quantity at that particular instant. From the above diagram, it is clear that the vertical component of the phasor is βAm sin(ππ‘ )β which is the instantaneous value of the quantity at instant βππ‘ β. β’ The speed of rotation of the phasor is equal to w rad/sec where π = 2Οf. β’ One rotation of the phasor corresponds to one cycle of the alternating waveform as shown in figure.
- 21. Numerical Q.1. An alternating current is given by π = πππ. ππππππππ.Find:(1) The maximum value (2) frequency (3) Time period (4) The instantaneous value when π is πππ. Solution: Given equation: π = 141.4π ππ314t comparing given equation of alternating current with standard form,π = πΌππ ππππ‘ We get, πΌπ = 141.4 , π = 314 1) Maximum value π°π = πππ. π π΄πππ 2) Frequency , π = 2ππ β΄ π = π 2π = 314 2π = ππ π―π 3) Time period, T = 1 π β΄ π = 1 50 = π. ππ πππ 4) Instantaneous value when π‘ = 3ππ π = 141.4π ππ314π‘ = 141.4π ππ314 β 3 β 10β3 β¦ β¦ β¦ . . (πππ‘π: π’π π π πππππ‘ππππ ππππ’πππ‘ππ ππ π ππ ππππ) π = πππ. ππ π¨πππ
- 22. Numerical Q.2. Express: i. π = ππβ πππ in rectangular form ii. π = ππ + ππ in polar form Solution: Using scientific calculator, i. π = 10β 600 = π + π£π. ππ β¦ ii. π = 16 + π8 = ππ. ππβ ππ. πππ β¦
- 23. Numerical Q.3. calculate frequency ,rms value, average value and amplitude of the waveform shown in fig. οSolution: from fig. we have ππππ ππππππ , π = 20ππ 1. πππππ’ππππ¦, π = 1 π β΄ π = 1 20β10β3 = ππ π―π 2. π΄πππππ‘π’ππ = ππππ π£πππ’π = πππ π½ π½π 3. π πππ = π = ππ 2 = 100 2 = ππ. ππ πππππ. 4. π ππ£πππππ = 2ππ π = 0.637 β π π = ππ. π π½ 100 π π 20ππ